Android根据资源名获取资源ID开发者社区

Android根据资源名获取资源ID

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Resources res = getResources(); final String packageName = getPackageName(); int imageResId = res.getIdentifier("ic_launcher", "drawable", packageName); int imageResIdByAnotherForm = res.getIdentifier(packageName + ":drawable/ic_launcher", null, null); int musicResId = res.getIdentifier("test", "raw", packageName); int notFoundResId = res.getIdentifier("activity_main", "drawable", packageName); Log.i(LOGTAG, "testGetResourceIds imageResId = " + imageResId + ";imageResIdByAnotherForm = " + imageResIdByAnotherForm + ";musicResId=" + musicResId + ";notFoundResId =" + notFoundResId);

1	I/MainActivity( 4537): testGetResourceIds imageResId = 2130837504;imageResIdByAnotherForm = 2130837504;musicResId=2130968576;notFoundResId =0

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/** * Return a resource identifier for the given resource name. A fully * qualified resource name is of the form "package:type/entry". The first * two components (package and type) are optional if defType and * defPackage, respectively, are specified here. * <p>Note: use of this function is discouraged. It is much more * efficient to retrieve resources by identifier than by name. * @param name The name of the desired resource. * @param defType Optional default resource type to find, if "type/" is * not included in the name. Can be null to require an * explicit type. * @param defPackage Optional default package to find, if "package:" is * not included in the name. Can be null to require an * explicit package. * @return int The associated resource identifier. Returns 0 if no such * resource was found. (0 is not a valid resource ID.) public int getIdentifier(String name, String defType, String defPackage) { try { return Integer.parseInt(name); } catch (Exception e) { // Ignore return mAssets.getResourceIdentifier(name, defType, defPackage); }

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Android根据资源名获取资源ID

Android根据资源名获取资源ID

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